Stats without TearsSolutions for Chapter 5

View orPrint:
These pages change
automatically for your screen or printer.
Underlined text, printed
URLs, and the table of contents become live links on screen;
and you can use your browser’s commands to change the size of
the text or search for key words.If you print, I suggest black-and-white,
two-sided printing.

Problem Set 1

1
(a) There are three coins, and each has two possible outcomes, so the
sample space will have 2³ = 8 entries.

(b)

S = {

HHH HTH THH TTH

}

HHT HTT THT TTT

(c) Three events out of eight equally likely events:
P(2H) = 3/8

Common mistake:
Sometimes students write the sample space correctly but
miss one of the combinations of 2 heads. I wish I could
offer some “magic bullet” for counting correctly, but the only advice
I have is just to be really careful.

2

Service type

Prob.

Landline and cell

58.2%

Landline only

37.4%

Cell only

2.8%

No phone

1.6%

Total

100.0%

(a) In a probability model, the probabilities must add to 1
(= 100%). The given probabilities add to 62.6%. What is the
missing 37.4%? They’ve accounted for cell and landline, cell
only, and nothing; the remaining possibility is landline only. The
model is shown at right.

(b)
P(Landline) = P(Landline only) + P(Landline and cell)

P(Landline) = 37.4% + 58.2% = 95.6%

Remark:
“Landline” and “cell” are not disjoint events,
because a given household could have both.
But “landline only” and “landline and cell” are
disjoint, because a given house can’t both have a cell phone
with landline and have no cell phone with landline.

3No, because the events are not disjoint.The figures are for being struck or
attacked, not killed. You’d have to be pretty unlucky to be
struck by lightning and attacked by a shark in the same year, but it
could happen.
If the question were about being killed by lightning or
by a shark, then the events would be disjoint and you could add the
probabilities.

4
(a) P(not A) = 1−P(A) = 1−0.7
→ P(not A) = 0.3

(b) That A and B are complementary means that one or the other
must happen, but not both. Therefore P(B) = P(not A)
→ P(B) = 0.3

(c) Since the events are complementary, they can’t both happen:
P(A and B) = 0

Common mistake:
Many students get (c) wrong, giving an answer of 1. If events
are complementary, they can’t both happen at the same time. That means
P(A and B) must be 0, the probability of something
impossible.

Maybe those students were thinking of P(A or B).
If A and B are complementary, then one or the other must happen, so
P(A or B) =
P(A) + P(B) = 1. But part (c) was about probability
and, not probability or.

5Yes, because the events are disjoint or mutually exclusive:
a person might have both cancer and heart disease, but the death
certificate will list one cause of death.
(1/5 + 1/7 ≈ 34%.)

6
P(divorced | man) is the probability that a randomly selected man
is divorced, or the proportion of men who are divorced.
P(man | divorced) is the probability that a randomly selected
divorced person is a man, or
the proportion of divorced persons that are men.

7
If the probability of a future event is zero, then that event
is impossible. If the probability of a past event is zero,
that just means that it didn’t happen in the cases that were
studied, not that it couldn’t have
happened.

This is the difference between theoretical and
empirical probability. A truly impossible event has a theoretical
probability of zero. But the 0 out of 412 figure is an empirical
probability (based on past
experience). Empirical
probabilities are just estimates of the “real” theoretical
probability. From the empirical 0/412, you can tell that the
theoretical probability is very low, but not necessarily zero. In
plain language, an unresolved complaint is unlikely, but just because
it hasn’t happened yet doesn’t mean it
can’t happen.

813/52 or 1/4

Common mistake:
Students often try some sort of complicated calculation here.
You would have to do that if conditions were stated on all five of
those cards, but they weren’t.
Think about it: any card has a
1/4 chance of being a spade.

9
S = { HH, HT, TH, TT }
(a) Three outcomes (HH, HT, TH) have at least one head. One of
the three has both coins heads. Therefore the probability is
1/3.
(b) Two outcomes (HH, HT) have heads on the first coin. One of the
two has both coins heads. Therefore the probability is
1/2.

10

(a)
0.0171 × 0.0171 =
0.0003

(b)
The events are not independent. When a married couple
are at home together or out together, any attack that involves one of
them will involve the other also.

11
(a) P(divorced) =
22.8/219.7 ≈ 0.1038

(b)
About 10.38% of American adults in 2006 were divorced. If you randomly selected an American adult in 2006, there was a 0.1038 probability that he or she was divorced.

(c)
Empirical or experimental

(d)
P(divorcedC) = 1−P(divorced) =
1−22.8/219.7 ≈ 0.8962About 89.62% of American adults in 2006 were not divorced
(or, had a marital status other than divorced).

(e)
P(man and married) = 63.6/219.7 ≈
0.2895 (You can’t use a formula on this
one.)

(f)
Add up P(man) and P(not man but married):

P(man or married) =
106.2/219.7 + 64.1/219.7 ≈
0.7751

Alternative solution:
By formula:

P(man or married) = P(man) +
P(married) − P(man and married)

P(man or married) = 106.2/219.7 +
127.7/219.7 − 63.6/219.7 =
0.7751

Remember, math “or” means one or the
other or both.

(g)
What proportion of males were never married?
30.3/106.2 = 28.53%.

(h)
P(man | married) uses the sub-subgroup of men within the subgroup of
married persons.

P(man | married) = 63.6/127.7 = 0.4980

49.80% of married persons were men.

Remark: You might be surprised that it’s under
50%. Isn’t polygamy illegal in the US? Yes, it is. But the
table considers only resident adults. Women tend to marry slightly
earlier than men, so fewer grooms than brides are under 18. Also,
soldiers deployed abroad are more likely to be male.

(i)
P(married | man) used the sub-subgroup of married persons within the
subgroup of men.

P(married | men) = 63.6/106.2 = 0.5989

59.89% of men were married.

12
P(five cards, all diamonds) = (13/52) × (12/51) ×
(11/50) × (10/49) × (9/48) ≈
0.0005(I was surprised that the probability is that high, about once every
2000 hands. And the probability of being dealt a five-card flush of
any suit is four times that, about once in every 500 hands.)

13

(a)
3 of 20 M&Ms are yellow, so 17 are not yellow. You want
the probability of three non-yellows in a row:
(17/20)×(16/19)×(15/18) ≈
0.5965

(b)
The probability is zero, since there are only two reds to
start with.

14
You’re being asked about all three possibilities: two fail,
one fails, none fail. Therefore the three probabilities must add up
to 1, and you need to compute only two of them.
It’s also important to note that the companies are
independent: whether one fails has nothing to do with whether the
other fails. (Without knowing that the companies are independent, you
could not compute the probability that both fail.)

(a) Since the companies are independent, you can use the simple
multiplication rule:

P(A bankrupt and W bankrupt) = P(A bankrupt) × P(W bankrupt)

P(A bankrupt and W bankrupt) = .9 × .8 = 0.72

At this point you could compute (b), but it’s
little messy because you need the probability that A fails and W is
okay, plus the probability that A is okay and W fails. (c) looks
easier, so do that first.

(c) “Neither bankrupt” means both are okay. Again,
the events are independent so you can use the simple multiplication
rule.

P(neither bankrupt) = P(A okay and W okay)

P(A okay) = 1−.9 = 0.1; P(W okay) = 1−.8 =
0.2

P(neither bankrupt) = .1 × .2 = 0.02

(b) is now a piece of cake.

P(only one bankrupt) = 1 − P(both bankrupt) − P(none bankrupt)

P(only one bankrupt) = 1 − .72 − .02 =
0.26

Remark:
If you have time, it’s always good to check your work
and work out (b) the long way.
You have only independent events (whether A is okay or fails, whether W
is okay or fails) and disjoint events (A fails and W okay, A okay and
W fails). The “okay” probabilities were computed in part
(c).

Common mistake:
When working this out the long way, students often solve only
half the problem. But when you have probability of exactly one out of
two, you have to consider both A-and-not-W and W-and-not-A.

You can’t use the “or” formula
here, even if you studied it.
That computes the probability of one or the other or both, but
you need the probability of one or the other but not both.

Remark:
If you computed all three probabilities the long way, pause a
moment to check your work by adding them to make sure you get 1.
Whenever possible,
check your work with a second type of computation.

(a)
15(You can assume independence because it’s a small sample
from a large population.)
P(red1 and red2 and
red3) = 0.13×0.13×0.13 = 0.0022

Common mistake:
Students sometimes compute 1−.13³. But
.13³ is the probability that all three are red, so
1−.13³ is the probability that fewer than three
(0, 1, or 2) are red. You need the probability that zero are red, not
the probability that 0, 1, or 2 are red.
Think carefully about where your “not” condition must be applied!

(c)
The complement is your friend with
“at least” problems. The complement of “at least one is
green” is “none of them is green”, which is the same as
“every one is something other than green.”P(green) = 0.16, P(non-green) = 1−0.16 = 0.84.
P(≥1 green of 3) = 1 − P(0 green of 3) = 1 −
P(3 non-green of 3) = 1−0.84³ ≈
0.4073

(d)
(Sequences are the most practical way to solve this
one.)
(A) G1 and G2C and G3C;
(B) G1C and G2 and G3C;
(C) G1C and G2C and G3
.16×(1−.16)×(1−.16) +
(1−.16)×.16×(1−.16) +
(1−.16)×(1−.16)×.16 ≈
0.3387

16
In “at least” and “no more than”
probability problems, the complement is often your friend. The
complement of “at least one had not
attended” is “all
had attended”. If the fans are randomly selected, their
attendance is
independent and you can use the simple multiplication rule.

P(all 5 attended) = 0.45^5 = 0.0185

P(at least 1 had not attended) = 1 − 0.0185 =
0.9815

17
Sequences are the way to go here:

(cherry1 and orange2) or
(orange1 and cherry2)

Common mistake:
There are two ways to get one of each: cherry followed by
orange and orange followed by cherry. You have to consider both
probabilities.

There are 11+9 = 20 sourballs in all, and Grace
is choosing the sourballs without replacement (one would hope!), so
the probabilities are:

(11/20)×(9/19) + (9/20)×(11/19) =
99/190 or about 0.5211

18The complement is your friend,
and the complement of “win at
least once in 5 years” is “win 0 times in 5
years” or “lose 5 times in 5 years”.

P(win ≥1) = 1−P(win 0) =
1−P(lose 5).

P(lose) = 1−P(win) = 1−(1/500) = 499/500

P(lose 5) = [P(lose)]5 = (499/500)^5 = 0.9900

P(win ≥1) = 1−P(lose 5) = 1−0.9900 =
0.0100 or 1.00%

Common mistake:
If you compute 1−(499/500)5 in
one step and get 0.00996008, be careful with your
rounding!
0.00996... rounds to 0.0100 or 1%, not 0.0010
or 0.1%.

Common mistake:
1/500 + 1/500 + ... is wrong. You can add
probabilities only when events are disjoint, and wins
in the various years are not disjoint events. It is possible (however
unlikely) to win more than once; otherwise it would make no sense for
the problem to talk about winning “at least once”.

Common mistake:
You can’t multiply 5 by anything.
Take an analogy: the probability of heads in
one coin flip is 50%. Does that mean that the probability of
heads in four flips is 4×50% = 200%? Obviously not!
Any process that leads to a probability >1 must be incorrect.

Common mistake:
1−(1/500)5 is wrong.
(1/500)5 is the probability of winning five years in a row,
so 1−(1/500)5 is the probability of winning 0 to 4
times. What the problem asks is the probability of winning 1 to 5
times.

19
(a), (b), and (c) are all the possibilities there are, so the
probabilities must total 1. You can compute two of them and then
subtract from 1 to get the third.

A very common mistake on problems like this is writing
down only one of the sequences. When you have exactly one success (or
exactly any definite number), almost always there are
multiple ways to get to that outcome.

You can’t use the “or” formula
here, even if you studied it.
That computes the probability of one or the other or both, but
you need the probability of one or the other but not both.

Problem Set 2

20
(a) P(ticket on route A) = P(taking route A) × P(speed trap on
route A) = 0.2×0.4 = 0.08. In the same way, the probabilities of
getting a ticket on routes B, C, D are 0.1×0.3 = 0.03,
0.5×0.2 = 0.10, and 0.2×0.3 = 0.06. He
can’t take more than one route to work on a given day, so those
are disjoint events. The probability that he gets a ticket on any one
morning is therefore 0.08+0.03+0.10+0.06 = 0.27.

(b) The probability of not getting a ticket on a given
morning is 1−0.27 = 0.73. The probability of getting no
tickets on five mornings in a row is therefore
0.735 ≈ 0.2073 or about 21%.

21
Two events A and B are independent if P(A|B) = P(A).

P(man) =
106.2/219.7 ≈0.4834

P(man|divorced) = 9.7/22.8 ≈
0.4254

Since P(man|divorced) ≠ P(man), the events are not
independent.

Alternative solution: You could equally well show that
P(divorced|man) ≠ P(divorced):

P(divorced|man) = 9.7/106.2 ≈
0.0913

P(divorced) = 22.8/219.7 ≈
0.1038

22
What’s the probability of ten of the same flip in a row? In
other words, given either result, what’s the
probability that the next nine will be the same? That must be
(1/2)9 = 1/512. You therefore expect this to happen
about once in about every 500 flips, or about twice in every thousand.